A-Level Chemistry OCR Notes

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Infrared Spectrometry

A pair of atoms joined by a chemical bond constantly vibrate. Heavy atoms will cause the bond to vibrate more slowly, and stronger bonds will vibrate more quickly.
The absorption of infra-red radiation increases the vibrational energy of a bond, by either bending or stretching

The degree of vibration depends on bond length, bond strength and the mass of the atoms joined
Particular bonds, and therefore functional groups, can be identified by looking at the frequencies absorbed in a infrared spectrum.
An infra-red spectrum is produced by a spectrometer:
- A beam of IR radiation in the range 200 – 4000 cm-1 passes through a sample
- The molecule absorbs some of the frequencies, and the emerging beam is analysed to identify the frequencies that have been absorbed
- A computer plots a graph of transmittance against wavenumber
Each peak in the spectrum represents the absorption by a specific bond vibration in the molecule
The bonds present in greenhouse gases (carbon dioxide, methane & water) can absorb infrared radiation that is reflected by the Earth’s surface, contributing to global warming.

Mass Spectrometry

A molecular ion (M+) is formed when molecules in a sample are ionised.
The molecular ion peak (the peak with the highest m/z ratio in the mass spectrum) can be used to determine the relative molecular mass
Molecular ions are able to break up into fragments during ionisation. This creates a fragmentation pattern on the mass spectrum. Different compounds will produce different fragment peaks

CH3OH → CH3OH+ + e- → CH3+ + OH•
Molecular ion fragment ion

Combined Techniques

The wide variety of analytical techniques that we have come across can be used to provide data and information to help us determine the structure of unknown compounds.
The molecular formula of a compound can be determined using the empirical formula and relative molecular mass of the molecule.
E.g. Determine the molecular formula of a compound with empirical formula CH2 and a relative molecular mass of 224. Relative molecular mass of the empirical formula:

C H2
12 + (1 × 2) = 14

Divide the relative molecular mass by that of the empirical formula:

224/14=16
Molecular formula:
16 x Ch2 = C16H32

Empirical & Molecular Formula

The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.
The empirical formula can be calculated from the composition by mass or percentage by mass. e.g 6.2 g of P is combined with O2 to form 14.2 g of phosphorous oxide. Calculate the empirical formula of the compound.

mass of O2: 14.2 g - 6.2 g = 8 g

number of moles of each element:

Divide through by the smallest number of moles to get the whole number ratio:

The molecular formula gives the number and type of atoms of each element in a molecule. It is made up of a whole number of empirical units.
The molecular formula can be determined using the empirical formula and relative molecular mass of the molecule. e.g. Determine the molecular formula of a compound with empirical formula CH2 and a relative molecular mass of 224.

Relative molecular mass of the empirical formula:

Empirical formula: P2O5

C H2
12 + (1 × 2) = 14

Divide the relative molecular mass by that of the empirical formula:
224/14=16

Molecular formula:
16 x CH2 = C16H32

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